Problem: Simplify and expand the following expression: $ \dfrac{1}{q - 9}+ \dfrac{5}{4q + 32}- \dfrac{3}{q^2 - q - 72} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{5}{4q + 32} = \dfrac{5}{4(q + 8)}$ We can factor the quadratic in the third term: $ \dfrac{3}{q^2 - q - 72} = \dfrac{3}{(q - 9)(q + 8)}$ Now we have: $ \dfrac{1}{q - 9}+ \dfrac{5}{4(q + 8)}- \dfrac{3}{(q - 9)(q + 8)} $ The least common multiple of the denominators is: $ (q - 9)(q + 8)$ In order to get the first term over $(q - 9)(q + 8)$ , multiply by $\dfrac{4(q + 8)}{4(q + 8)}$ $ \dfrac{1}{q - 9} \times \dfrac{4(q + 8)}{4(q + 8)} = \dfrac{4(q + 8)}{(q - 9)(q + 8)} $ In order to get the second term over $(q - 9)(q + 8)$ , multiply by $\dfrac{q - 9}{q - 9}$ $ \dfrac{5}{4(q + 8)} \times \dfrac{q - 9}{q - 9} = \dfrac{5(q - 9)}{(q - 9)(q + 8)} $ In order to get the third term over $(q - 9)(q + 8)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{3}{(q - 9)(q + 8)} \times \dfrac{4}{4} = \dfrac{12}{(q - 9)(q + 8)} $ Now we have: $ \dfrac{4(q + 8)}{(q - 9)(q + 8)} + \dfrac{5(q - 9)}{(q - 9)(q + 8)} - \dfrac{12}{(q - 9)(q + 8)} $ $ = \dfrac{ 4(q + 8) + 5(q - 9) - 12} {(q - 9)(q + 8)} $ Expand: $ = \dfrac{4q + 32 + 5q - 45 - 12}{4q^2 - 4q - 288} $ $ = \dfrac{9q - 25}{4q^2 - 4q - 288}$